Free body diagram of a wheel driven by a motor

[EM_Guy]

How about we have that discussion after we agree on the right equation for the FBD? You seem to be acknowledging that as the angular speed of the wheel increases beyond a certain point, even the spokes on the bottom of the wheel could be in tension. But when we draw a FBD of the part of the wheel touching the ground, you include no tension in your diagram or in your equation.

 

[A.T.]

How about we have that discussion after we agree on the right equation for the FBD?

Well, draw one. Define the bodies and the interactions between them.

 

[EM_Guy]

See post #27 and my correction in post #29. There is the equation describing the FBD of the part of the wheel touching the ground. In post #28, you did not include a tension force.

 

[A.T.]

There is the equation describing the FBD of the part of the wheel touching the ground.

A FBD defines interacting bodies and labels the forces between them. Which terms correspond to which body exerting a force on the part of the wheel touching the ground?

 

[EM_Guy]

From post #29:

-Mg = downward force of vehicle exerted through the spoke onto the section of wheel touching the ground
-mg = weight of section of wheel touching the ground
N = normal upward force of ground acting on the section of wheel touching the ground
T = tension in spoke attaching the section of wheel touching the ground to the center of the wheel

 

[A.T.]

-Mg = downward force of vehicle exerted through the spoke onto the section of wheel touching the ground
T = tension in spoke attaching the section of wheel touching the ground to the center of the wheel

If the spoke is one body, then there is one force F_{spoke on section} exerted by the spoke on the section of wheel touching the ground. And that force is either downward or upward or zero.

 

[EM_Guy]

If the spoke is one body, then there is one force F_{spoke on section} exerted by the spoke on the section of wheel touching the ground. And that force is either downward or upward or zero.

There is one net force exerted by the spoke on the section of wheel touching the ground. It will be downward, upward, or zero. But that net force can be described as the superposition of two forces (as I have described).

 

[A.T.]

But that net force can be described as the superposition of two forces

You can just as well decompose the ground reaction force into two forces, one of which equals m*a, and claim that this is the force that "causes acceleration". It's arbitrary and meaningless.

 

[EM_Guy]

You can just as well decompose the ground reaction force into two forces, one of which equals m*a, and claim that this is the force that "causes acceleration". It's arbitrary and meaningless.

I believe you are off base here. But let me suspend that belief for a second and assume that you are correct.

I have said: -Mg - mg + N + T = ma = mv^2/r

You have said: -Mg -mg + N = ma

I have also said: -Mg - mg + N = 0 (Consider a FBD of the entire vehicle. The vehicle isn't lifting up off of the earth.)

You have not replied to this. Let me ask you point blank: Do you or do you not agree that this equation describes the FBD of the vehicle?
-Mg - mg + N = (M+m)ay

ay = acceleration in the y-direction.
M is a constant (the mass of vehicle - less the section of wheel touching the ground).
m is a constant (the mass of the section of wheel touching the ground).
g is obviously a constant.

Let's put it this way, I'm saying that the net upward force that the spoke exerts on the section of wheel touching the ground is this:

Fspoke = T - Mg

You are saying that the net upward force that the spoke exerts on the section of wheel touching the ground is this:

Fspoke = -Mg

Now, neither M nor g change as the angular speed of the wheel changes, so you are telling me then that the angular speed of the vehicle does not impact one bit the force of the spoke on the section of wheel touching the ground.

Yet, you also say:

if the wheel spins fast enough, even the lowest spoke might come under tension

Do you not think that you are contradicting yourself?

You certainly make assertions with confidence. Unless you are right and confident that you are right, you should not express yourself with so much confidence.

 

[A.T.]

You have said: -Mg -mg + N = ma

Ok, I see what you mean. I reused "-Mg" for "the total force from all other wheel parts (spoke and circumference) acting on the piece of wheel touching the ground." which is wrong.

I have said: -Mg - mg + N + T = ma = mv^2/r

This is basically decomposing F_{spoke on section} into -Mg and ma. Where you loose me is calling ma "spoke tension" (T) and claiming it is "causing the acceleration".

 

[EM_Guy]

This is basically decomposing F_{spoke on section} into -Mg and ma. Where you loose me is calling ma "spoke tension" (T) and claiming it is "causing the acceleration".

Okay, let's start at the beginning. We just have a wheel. The axle is unloaded. No vehicle. No gravity even. Just a wheel in space outside of any gravitational field.

Let us examine the force acting on one little section of the rim that is connected to a spoke.

If the wheel is not spinning, then the net force acting on that section of the rim is zero. We have no gravity, no normal force, no tension or compression in the spoke.

Okay, now, let's say that the wheel is spinning at some constant angular speed.

Now, if we consider a FBD of a small section of the wheel, we have this:

Fspoke = ma.

And we know that a = v^2/r from kinematic constraints. Fspoke = tension = T.

Okay...

Now, scenario two: We have a wheel on a unicycle on the Earth and a person sitting on the unicycle. No rolling, no motion.

FBD of small section of the wheel touching the ground:

Fspoke - mg + N = 0.

Fspoke = -Mg. The spoke is in compression. The weight of the unicycle and person sitting on the unicycle is transmitted through the spoke pushing the section of wheel on the ground down. The normal force is pushing it up.

Okay, now let the wheel roll at some angular speed.

Fspoke - mg + N = ma. (This is the equation corresponding to the FBD of the section of the wheel that is touching the ground).

But now Fspoke is the superposition of the weight of the unicycle and person sitting on the unicycle (causing compression) and the tension that arises due to the fact that you have a piece of a wheel connected to the rest of the wheel (which we are taking as a rigid body). Specifically, the piece of the wheel touching the ground is connected to the center of the wheel through the spoke.

Thus, Fspoke = T - Mg. (I am using superposition here. Mg due to the weight of the unicycle / rider compressing the spoke, while the angular speed of a rigid body causes tension in the spoke. Is the spoke under net tension or net compression? It depends how fast the wheel is spinning and the mass of the system).

Therefore,

T - Mg - mg + N = ma. (This is the equation corresponding to the FBD of the section of the wheel that is touching the ground - keeping in mind that I have done a simple substitution here for Fspoke).

Again, a = v^2/r.

At the same time, the unicycle and the rider are not accelerating upwards off the ground. This is the crux of my argument. So if we draw a FBD of the whole system, we get:

-Mg - mg + N = 0. (This is the equation corresponding to the FBD of the whole unicycle / rider system.)

Note that while the section of wheel that is touching the ground is accelerating upward, the system taken as a whole is not accelerating upward.

Therefore,

T = ma. (This is the equation corresponding to the FBD of the section of wheel that is touching the ground - simplified - given the equation corresponding to the FBD of the whole unicycle / rider system).

Note: This does not imply that the spoke is under any net tension.

a = v^2 / r.

Now, let's consider the compressive and tension stress in the spoke. Let negative correspond with compression and positive correspond with tension:

(T - Mg)/A = net tension stress.

But I have said that T = ma = mv^2/r.

So,

(mv^2/r - Mg) / A = net tension stress.

Therefore, if mv^2/r > Mg, the spoke is under positive net tension. If mv^2/r < Mg, the spoke is under positive net compression (aka negative net tension). If mv^2/r = Mg, the spoke is under no stress at all.

 

[EM_Guy]

The crux of my argument is that we can (and we do) have a situation in which forces are simultaneously trying to place the spoke under compression and tension.

Imagine a rigid box. You can imagine simultaneously compression forces applied to the box - trying the squeeze the box, while tension forces are trying to stretch the box. Is the box under compression or tension?

 

[A.T.]

Okay, let's start at the beginning. We just have a wheel. The axle is unloaded. No vehicle. No gravity even. Just a wheel in space outside of any gravitational field.

When you have only a single force acting on the section (like when spinning in space), then I can see how you can say that this force "causes the acceleration". But as soon you have multiple forces, then I don't see the physical significance of such attribution. You can represent any of the forces as ma and the remainder.

The crux of my argument is that we can (and we do) have a situation in which forces are simultaneously trying to place the spoke under compression and tension.

It's not a situation. It just the way you have chosen to decompose two compressive forces (for slow rotation).

Imagine a rigid box. You can imagine simultaneously compression forces applied to the box - trying the squeeze the box, while tension forces are trying to stretch the box. Is the box under compression or tension?

Let's say you push the box with 2N from left and push 1N from right. It's obviously under compression and accelerating to the right. But then you choose to represent the 1N right push force as a superposition of 2N push and 1N pull, then you call the 1N pull is a "tensional component" which "causes the acceleration". That's what it all boils down to, as far I can see.

 

[EM_Guy]

You are getting distracted by my box analogy. Go back to post # 41. Which, if any, of my equations are wrong and why?

 

[A.T.]

You are getting distracted by my box analogy. Go back to post # 41. Which, if any, of my equations are wrong and why?

It's not the equations, but your interpretation of them. The simple box example demonstrates its arbitrariness very well.

 

[EM_Guy]

So my equations are right, but my interpretations are wrong? Which interpretations are wrong?

You missed the point of the box analogy. If the box is under compression then you have two forces acting on it - squeezing it. The net force is thus zero, and the box doesn't accelerate, but it is under compression. If the box is under tension, then you have two forces acting on it - trying to pull it apart. The box doesn't accelerate, but the box is under tension.

Bottom line, if you are not going to take the time to go step by step through post #41, then I do not see that it will benefit either of us to continue this discussion.

 

[A.T.]

Which interpretations are wrong?

Not wrong, but arbitrary: The attribution of the total acceleration to some component of one of multiple forces.

If the box is under compression then you have two forces acting on it - squeezing it. The net force is thus zero, and the box doesn't accelerate,

Compression and acceleration aren't mutually exclusive, and since our wheel section is accelerating so should the box in the analogy.

 

[A.T.]

f mv^2/r = Mg, the spoke is under no stress at all.

Note that in this case the spoke exerts no force on the bottom wheel section. So the only force acting on the bottom wheel section is the force from the ground. And yet you still choose to attribute the acceleration not to that single force acting from the ground, but to some "tension component" of a spoke that isn't under tension and exerts no force?

 

[EM_Guy]

Not wrong, but arbitrary: The attribution of the total acceleration to some component of one of multiple forces.

It is not arbitrary. I have two FBDs - one for the entire unicycle / person system and one for the piece of wheel that is touching the ground. The two equations share terms.

And yet you still choose to attribute the acceleration not to that single force acting from the ground, but to some "tension component" of a spoke that isn't under tension and exerts no force?

T - Mg - mg + N = ma

Fspoke = T - Mg

So, you can say

Fspoke - mg + N = ma

And if Fspoke = 0, then

-mg + N = ma

Granted. When there is no stress in the spoke, N - mg causes the upward acceleration of the segment of the wheel touching the ground. This is true.

But what I have been saying is true too.

-Mg - mg + N = 0

Therefore, T = ma.

This tension isn't a "net tension" but a "tension component" in the Fspoke = T - Mg vector. And it is not arbitrary; it is deduced from the FBD of the entire unicycle / person system.

 

[jbriggs444]

Okay, let's start at the beginning. We just have a wheel. The axle is unloaded. No vehicle. No gravity even. Just a wheel in space outside of any gravitational field.

Let us examine the force acting on one little section of the rim that is connected to a spoke.

If the wheel is not spinning, then the net force acting on that section of the rim is zero. We have no gravity, no normal force, no tension or compression in the spoke.

Stop right there. That claim is unsupported. The rim may be under tension or compression.

 

[EM_Guy]

Stop right there. That claim is unsupported. The rim may be under tension or compression.

This depends on how the wheel is designed and fabricated, no? For the purposes of this exercise, I'd like us to assume that the spokes are neither in compression or tension.

 

[jbriggs444]

This depends on how the wheel is designed and fabricated, no? For the purposes of this exercise, I'd like us to assume that the spokes are neither in compression or tension.

So you are contemplating a design like that seen here? http://www.strangevehicles.com/content/item/147049.html

 

[EM_Guy]

So you are contemplating a design like that seen here? http://www.strangevehicles.com/content/item/147049.html

I was not. Is it inconceivable to have a wheel whose spokes are neither in compression nor tension?

 

[jbriggs444]

I was not. Is it inconceivable to have a wheel whose spokes are neither in compression nor tension?

It is inconceivable to have a wheel where the assertion is that the only relevant force is from the spokes and the ground if the wheel also has a rim.

Edit: Let me try to take some spin off that remark. Upthread we have references to forces on small sections of rim. For that to make sense, we have to de-couple sections of rim from one another. How better to think of that than a collection of shoes on spokes with no rim?

Edit: If we hypothesize a rim that is neither in tension nor compression and a collection of spokes that are neither in tension nor compression and we spin that wheel up so that there is now a centripetal force, we have a problem. The system is over-determined. There is no way to know whether the spokes are supplying the centripetal force or whether the rim is doing so. Eliminating the rim eliminates that over-determination.

 

[EM_Guy]

It is inconceivable to have a wheel where the assertion is that the only relevant force is from the spokes and the ground if the wheel also has a rim.

Okay. So, can you draw (or describe) a FBD of the section of wheel touching the ground - given that it has a rim? As far as I see it, this would add two forces acting on the section of the wheel touching the ground (either compressive or tension forces) and that these forces would be equal in magnitude and opposite in direction, right? Thus, would they not cancel each other out? Do they need to be included in this analysis?

 

[jbriggs444]

Okay. So, can you draw (or describe) a FBD of the section of wheel touching the ground - given that it has a rim? As far as I see it, this would add two forces acting on the section of the wheel touching the ground (either compressive or tension forces) and that these forces would be equal in magnitude and opposite in direction, right? Thus, would they not cancel each other out? Do they need to be included in this analysis?

Excellent. Now we are making progress. I agree that this is a correct way of looking at the section of the wheel. And let us make the assumption of an ideal rim which supports compression and tension forces but not bending or shear.

Are the forces at the two ends really equal and opposite? Let's take the easier question first. Are they really opposite?

 

[EM_Guy]

Excellent. Now we are making progress. I agree that this is a correct way of looking at the section of the wheel. And let us make the assumption of an ideal rim which supports compression and tension forces but not shear.

Are the forces at the two ends really equal and opposite? Let's take the easier question first. Are they really opposite?

I'm considering the wheel touching the ground at a single point. I know that this is not realistic. But an ideal circular wheel would only touch the ground at a single point. The tension or compressive forces from the rim would therefore be tangential to that point - no?

Now, if the rim is spinning with some angular speed, then on one side of our point of interest, we have part of the rim with momentum toward the point of interest, and on the other side, we have momentum away from the point of interest.

Do the momenta of the particles of the rim affect that compressive / tension forces of the rim?

 

[jbriggs444]

I'm considering the wheel touching the ground at a single point. I know that this is not realistic. But an ideal circular wheel would only touch the ground at a single point. The tension or compressive forces from the rim would therefore be tangential to that point - no?

An ideal circular wheel with a spoke at every point no longer can no longer be considered to have spokes at any point. It is, instead, a disc.

Edit: So consider the spacing between your spokes. What about the forces on the section of rim centered on a particular spoke?

 

[EM_Guy]

An ideal circular wheel with a spoke at every point no longer can no longer be considered to have spokes at any point. It is, instead, a disc.

With respect, so what? 1. I never said that there is a spoke at every point. 2. I don't think it much matters whether we are dealing with a disk or a wheel with spokes.

But to make everything more simple, let's say that there are a finite number of spokes on the wheel. And they are spaced out equiangularly (if that is a word).

 

[jbriggs444]

With respect, so what? 1. I never said that there is a spoke at every point. 2. I don't think it much matters whether we are dealing with a disk or a wheel with spokes.

If there it not a spoke at every point then why would one assume that there is a spoke where the wheel touches the ground! You need one there to prevent the wheel from deforming, surely?

But to make everything more simple, let's say that there are a finite number of spokes on the wheel. And they are spaced out equiangularly (if that is a word).

So now are the two forces from rim tension or compression oriented in opposite directions?